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Trigonometry

Question 1 of 9

Area of Triangle

Question

Find the area of \(\triangle ABC\) whose \(c = 6.5\,\text{cm}\), \(b = 13\,\text{cm}\) and \(\angle A = 100^\circ\) (Round your answer to 2 decimal places)

Show solution Hide solution Fully worked — 3 steps
Triangle ABC: vertex A at the bottom left, vertex B above it and vertex C at the bottom right. The angle at A, between sides c and b, is marked 100 degrees. Side c, from A up to B, is 6.5 cm; side b, from A to C along the base, is 13 cm; side a, from B to C, is unlabelled. The solution uses the two given sides b and c and their included angle A.
  1. State the formula
    \[ \mathit{area} = \frac{1}{2} bc \sin A \]
  2. Substitute values based on formula
    \[ \mathit{area} = \frac{1}{2}(6.5\,\text{cm})(13\,\text{cm})(\sin 100^\circ) \]
  3. Final answer

    Evaluate \(\frac{1}{2}(6.5\,\text{cm})(13\,\text{cm})(\sin 100^\circ)\):

    \[ \mathit{area} = 41.61\,\text{cm}^2 \]

    That is, the area of triangle ABC is 41.61 square centimetres, to 2 decimal places.